Electronic – How to calculate voltage drops and power for a current dependent current source in series

current-sourcepowerseries

So the question goes as follows:
For the circuit in the figure below determine: a) the loop current, b) voltage drops across all the resistors; c) voltage of the dependent current source, d) power associated with each element and whether it is absorbed or delivered.

I don't need the entire answer done, just maybe the first step to get started.

Thanks to everyone that told me to do the KVL, but I'm stuck trying to express the dependent current source. I know it is 5 times the current of I, but this is where I get stuck. VR1 + VR2 – VIs + VR3 + E1 = 0

A series circuit with a voltage source (30V), two resistors (5 and 4 ohms respectively), a current dependent current source at 5*I, and another resistor (6 ohms). The current I is the current between the voltage source and 5 ohm resistor.

Best Answer

I think you are on the right track with your initial KVL. I will re-write it here:

\$ V_{R1} + V_{R2} - V_{Is} + V_{R3} + E_1 = 0\$

There is only one typo, which is the sign of voltage source \$E_1\$. Based on the direction of the \$I\$ arrow, the correct KVL would be:

\$ V_{R1} + V_{R2} - V_{Is} + V_{R3} - E_1 = 0\$

Since we know \$E_1=30\$, what we have is one equation in 4 unknowns. Since we only have one equation, we want to reduce this to a single unknown, to arrive at a unique (if it exists) solution. We can make use of two facts:

For a resistor, \$V=IR\$

Also, for the dependent current source, \$V_{Is} = 5I\$.

So in the end, using these two facts can reduce the above KVL to a single equation in \$I\$, which you can solve.

There is nothing wrong with proceeding in the above analysis, it's actually the best way to be correct. Over time you will start to see little shortcuts, which can simplify the equations. For example, in this circuit a current-dependent voltage source (CCVS) whose voltage is dependent on its own current is simply behaving as a resistor. Thus that CCVS has the same \$I/V\$ characteristics as a 5\$\Omega\$ resistor, so the loop current is the 30V dropped across a total of \$(5+4+5+6)=20\Omega\$.

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